Question

Estado de oxidacion de SO42-, NO2-, SiO32- y K2Cr2O7 urgenteee D:

Answer 1

oxidation state of SO42-, NO2-, SiO32- and K2Cr2O7

Answer:

2-

1-

2-

0

Explanation:

The oxidation states of the ions and compounds is given as;

SO₄²⁻  = 2-

NO₂⁻ = 1-

SiO₃²⁻  = 2-

K₂Cr₂O₇  = 0

Here are some rules that governs assignment of oxidation states;

1. For elements in uncombined states, the oxidation state is 0

2. For simple ions, the oxidation number is the actual charge carried.

3. The algebraic sum of all oxidation number of all atoms in a neutral compound is zero.